MCQ
Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three units vectors such that $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0} .$ If $\lambda=\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}} $ and $\overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}},$ then the ordered pair $(\lambda, {\mathrm{\vec d}})$ is equal to 
  • $\left(-\frac{3}{2}, 3 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\right)$
  • B
    $\left(-\frac{3}{2}, 3 \overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}\right)$
  • C
    $\left(\frac{3}{2}, 3 \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}\right)$
  • D
    $\left(\frac{3}{2}, 3 \overline{\mathfrak{a}} \times \overline{\mathfrak{c}}\right)$

Answer

Correct option: A.
$\left(-\frac{3}{2}, 3 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\right)$
a
$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0}$

$\Rightarrow|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+|\overrightarrow{\mathrm{c}}|^{2}+2(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}})+2(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}})+2(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})=0$

$\lambda=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=\frac{-3}{2}$

$\overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}$

$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0}$

$\Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}$

$\Rightarrow \overrightarrow{\mathrm{d}}=3(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})$

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